OURNAL
OF COMBINATORIAL
Graph
Series B 36, 16-25
THEORY,
Decompositions
(1984)
without
Isolates
NATHAN LINIAL*
University
of California,
Los Angeles,
Communicated
Received
California
90024
by the Editors
May
2;.
1981
A. Frank
(Problem
session of the Fifth
British
Combinatorial
Conference,
Aberdeen,
Scotland,
1975) conjectured
that if G = (V, E) is a connected graph with
all valencies
>k and a,,..., ak > 2 are integers with C ai = / VI, then k’ may be
decomposed
into subsets A, ,..,, A, so that lAil = ai and the subgraph spanned by Ai
in G has no isolated vertices (i= l,.... k). The case k = 2 is proved in Maurer
(J.
Combin.
Theory Ser. B 27 (1979)
294-3 19) along with some extensions.
The
conjecture
for k = 3 and a result stronger
than Maurer’s
extension
for k = 2 are
proved.
A related characterization
of a k-connected
graph is also included
in the
paper, and a proof of the conjecture
for the case a, = a2 = ... = ak-, = 2.
Graph theoretic terminology
is standard; see [ 1, 2] for definitions.
graph G = (I’, E) has order u = ] I’]. For A, B g V we let
EWB)=
A
{[x,ylEElxEA,yEB),
E(A)=W,A),
Also (/I& = (A) is the subgraph of G spanned by A. We let 6(G) be the
smallest valence of vertices in G.
In [3] A. Frank made the following:
Conjecture. Let G = (V, E) be a connected graph, with 6(G) > k. Let
a, ,..., ak > 2 be integers with Cf ai = v = 1VJ. Then V may be decomposed
into A , ,..., A, so that lAil = ai and (Ai) has no isolated vertices (i = l,..., k).
A graph for which the conclusion of the conjecture holds is said to be kdecomposable, so the conjecture says that a connected graph with 6 > k is kdecomposable.
* The author’s
author is grateful
work was supported
by the Chaim
for the grant’s generous support.
16
0095-8956184
Copyright
All rights
$3.00
CD 1984 by Academic
Press. Inc.
of reproduction
in any form reserved.
Weizman
Postdoctoral
Fellowship.
The
GRAPH DECOMPOSITIONS
17
This problem for k = 2 was discussed by Maurer [6] who also considers
the computational complexity of finding these, and related, decompositions
of graphs. Maurer proved Frank’s conjecture for k = 2 and proved some
extensions of this case. Among others he proves
THEOREM M [6]. Let G = (V, E) be a connected graph, 6(G) > 2. Let
a,, a, > 2 be integers with a, + a, = v = / VI. Then V may be decomposed
into A,, A, so that /Ai1 = ai (i = 1,2) one of the (Ai) is connected and the
other one has no isolated vertices.
The results of this paper are: a proof of the conjecture for k = 3
(Theorem 2), a theorem which contains Theorem M, a related characterization of k-connected graphs (Theorem 3), and a proof of the conjecture
fora,=a,=..-=a,-,=2.
Our first theorem contains Theorem M. To state it we define a friendship
U {xiIn>i>
1) U {yiIn>i>
l},
graph J’, = (V,, En), where V,={p}
E,={[xi,Yi]ln>i>l}
U ([p,Xi]ln>i>l}
U ([PpYi]In>i>l}.
In
other words F, = K, + nK,. Now we state
THEOREM
1. Let G = (V, E) be a connected graph, 6(G) > 2, and let
a,,a,>
2 be integers such that (VI = v >a, +a,. Then unless G is a
friendship graph and both a,, a, are odd, there exist A,, A, c V so that
A,nA,=#,
jA,l=a,
(i= 1,2), one of(A,) is connected and the other one
has no isolated vertices.
Proof
Note first that if G is a friendship graph and a,, a, are odd then
at least one of the (Ai) must have an isolated vertex. We need
LEMMA
1. Let G = (V, E) be a connected graph, 6(G) > 2, which is not
a friendship graph. Then there are two adjacent vertices x, y E V so that
4G.J > 2, where G,, is the graph obtained from G by contracting x, y to a
single vertex.
Proof: Let us consider the vertices of degree 23. If there are none G
must be a circuit and the lemma holds unless G = K, which is the friendship
graph F, . If there is just one vertex of degree 23, G is a collection of circuits
having exactly one vertex in common. For such graphs the conclusion of the
lemma holds except if all circuits are triangles and the graph is a friendship
graph.
If there are two adjacent vertices p, q with d(p), d(q) > 3, then either
6(G,,) > 2 and we let x =p, y = q, or 6(G,,) = 1. In the latter case there
must be a vertex w with T(w) = {p, q}. Let x =p, y = w to achieve
4G,,) > 2.
In the remaining case we can assume that there exist vertices p, q with
d(p), d(q) 2 3 and such vertices must be nonadjacent. Now let p = x0, x1 ,...,
18
NATHAN LINIAL
x, = q be a shortest path between them. We claim that for x = p, y = x, we
have 6(G,,) > 2. Otherwise p and x, must have a common neighbour, but
since d(x,) = 2, T(x,) = {p, xz} and if p. x2 E E, there is a shorter path from
P to 4.
We go back to prove the theorem by induction on c = v - (a, + a*). The
case c = 0 is Theorem M above. So assumec > 1 and G is not a friendship
graph. Find x, y which satisfy the lemma and consider G’ = G,,. If G’ is a
friendship graph then it is easy to check that the theorem holds. If G’ is not
a friendship graph we may apply induction:
Let p be the vertex in G’ which represents (x, y}. By the induction
hypothesis we may find disjoint subsetsAi, Ai of q{x,y} U (p) so that
lA\/=ai
(i= 1,2), one of (A;),, is connected and the other one has no
isolated vertices. If p & Ai U Ai, let A i = AI (i = 1. 2) and this satisfies the
theorem.
Assume, then, that p E Ai and let A; = A{\(p} U (x,y),
A, = A; be sets
of vertices in G. They fail to satisfy the theorem only in that IA; / = a, + 1.
Since p belongs to a component of (Ai)c of order 22, x, y belong to a
component of (A;)c of order 23. Omitting a non-cut vertex of this
component A, is obtained and the theorem follows. 1
Let us state and prove now the main result. We prove the conjecture for
k= 3.
THEOREM 2. Let G = (V, E) be a connected graph with all valencies 23.
Let a,, a,, a3 > 2 be integers with a, + a, + a3 = v = / I/). Then there is a
decomposition A,, A,, A, of V so that 1A i / = a, and (A i) has no isolated
vertices (i = 1, 2, 3).
First we need two technical lemmas:
LEMMA M. Let G be a graph with 6(G) > 2. If all components of G are
of order >5 then G is 2-decomposable.
Proof.
Contained in [6, Theorem 4.211.
LEMMA 2. Assume 6(G) > 3 implies 3-decomposability for connected
graphs G of order <v. Then it implies 3-decomposability also for graphs of
order <v having all componentsof order >6.
Proof. By induction on the order of the graph. Let c, > ‘.. > ck > 6 be
the orders of the components of G and let a, > a2 > a3 > 2 satisfy
a, + a, + a3 = v. If ck < a, - 2 we may continue by induction so assume
ck > a, - 1 which readily implies k < 3, and since k > 1 we have to check
only k = 2, 3.
Let k = 3 first. Of course c, < a, but in case of equality we may proceed
GRAPH
19
DECOMPOSITIONS
by induction. So we may assume c3 = a, - 1. Similarly cz > a, - 1 and
c2 # a, imply c2 = a2 - 1 (c, > a, + 1 is impossible since a, + a, + a3 =
c, > c, > c3). Using the same arguments we
c1+cz+c,,
a,>a,>a,,
remain with two cases:
C2
at3
a
a3
a
a
a+1
a+1
a+1
a
a
a+1
at1
a-2
with a > 6. Each of these can be handled easily and the details are omitted.
For k = 2 we find integers q, , q2, q3 with ai-2>q,>2,
or qi=aj
(i = 1, 2, 3) and C qi = c,. Then we decompose c, with parameters q,, q2, q3
This yields a solution unless c, = 7,
and cl with a, -ql,
a,-qq,, ax-q,.
CI>= a3 = 3. which can be easily handled. I
Proof of Theorem 2. First we show that G may be decomposed into
nontrivial stars. Namely, we want to find a set of vertices R = (r, ,..., r,} and
nonempty subsets L i ,..., L, of V with T(ri) ?L, (1 < i< m) so that R,
L i ,..., L, is a decomposition of I’.
Let R, L , ,,.., L, satisfy the above conditions except that R U (U’: L/) # V
and let IR U (Uy Li)l be largest possible. Since G is connected there is an
.Y E V\(R U (Uy L,)) with a neighbour in R U (UT Li). By maximality this
neighbour cannot be in R. If [X,JJ] E E, y E Li and lLil > 2 we let
Lf = Li\4’, rm+, =.V, L,+, = {x) contradicting the maximality. If Li = { y},
let r( = y, LI = {ri, X} again contradicting maximality.
Define Si=(ri}ULi.
si=ISil
(m>i>
1) and assume s,>...>s,.
Among all decompositions into starts (R, L, ,..., L,) choose one with largest
minimal.
m and among those, choose one with (s, ,..., s,) lexicographically
These assumptions imply
if
si > 4,
then
if
si + sj > 6, i #j,
e(L,) = 0;
then
(1)
e(L,, Lj) = 0.
(2)
Besides, if e(L,, rj) # 0, then sj > si - 1. We say that Sj can be reached from
Si if there is a sequence i = i,,..., i, =j (t > 0) without repetitions such that
e(LiL,, ri, ,) # 0 (v = O..... t - 1). The last observation extends to
if
Sj can be reached from Si,
then
si>si-
1.
In the following section we assume s, > 4. Consider now all stars S, ,..., S,
with si = s, (p > i > l), and let P = {m > i > 1 ISi can be reached from one
of s i,..., S,}. Let H= ((JicP Si). We claim that H is 3-decomposable.
Referring to Lemma 2 we note that all components of H have order >6. Also
all valencies in H are 23, this can fail for a vertex x E Lj (i E P) only if x
20
NATHAN
LINIAL
has a neighbour in UidP Li which by (2) is possible only if s, = 4, si = 3,
and [x, y] E E, y E Lj, sj = 2. But then we can replace a 4, 3, 2 subsequence
of s 1Y--*9s, by the lexicographically
smaller 3, 3, 3. For rj, j E P, the
condition dH(rj) 2 3 can fail only if sj = 3, but since s, > 4, the edge by
which Sj was reached from a larger star ensures that indeed dH(rj) > 3.
We want to reduce the proof to the case where P = (l,..., m}. If
that IP( > 3,
p f { l,..., m} let t = max{sj ]j@ P}. We already know
si > s, - 1 (i E P) and so CipP si > 3s, - 2.
If t<a,-2
we can replace a,, a2, a3, the parameters for decomposing,
by a1 - t, a*, a3, and move to the next largest Sj (j 66P). If this process can
be carried out until all stars not in P are used we finally have to 3decompose H with parameters a;, a;, a; > 2, which can be done by
induction on U. So consider the first case where it fails. Assume, then,
t > a, - 1 and use a, > a, > a3, s, > t + 1, xi,, si > 3s, - 2 to write
2s, + t + 1 > 3t + 3 > 3a, > a, + a, + a3 = u
~ t + ~~ Si ~ t + 3S, - 2
which implies 3 > s,, a contradiction.
This allows us to assumefrom now on that s, > s, - 1. Moreover we may
assumethat Uy Li is an independent set of vertices, if s, > 4. If s, > 5 this
follows immediately from (l), (2). If s, = 4, (2) reduces the discussion to a
case where some si = 3, Li = {x,y} and [x, y] E E. But since Si can be
reached from a star on 4 vertices we may transfer vertices and transform Si
to a star on 4 vertices which violates (1).
Besides, we are allowed to assume that e(rj, UT Li) > 3 (m >j> 1).
Again if s, > 5 this is clear and if s, = 4 and sj = 3, rj has a neighbour in
fJi+jLi, since Sj can be reached from other stars.
We claim that we may assume e(R) = 0. Otherwise start deleting edges
from E(R). On deleting such an edge all valencies in G remain >3 but it may
possibly disconnect. So assume that one of these edges is a bridge. By
Lemma 2 we may assumethat at least one of the components of the graph
resulting when this edge is deleted is of order <5. This leads to a short list of
possible cases
Sizes of Components
ala2a3
474
54
535
I,4
735
3,3,2
3,393
4,492
5,393
4,4,4
GRAPH
21
DECOMPOSITIONS
Each one of these may be handled separately.
So we may assume
d(x) = 3
(xEULi)*
Consider now the graphs Gi = G\( {ri} U T(ri)) (m > i > 1). We want to
show that each of them contains a vertex of valence <2. If 6(G,) > 3 for
some i, we claim that all components of Gi have order >6. This follows
easily, since G is bipartite and has all valencies >3. This means that
Lemma 2 will be applicable. Let q = d(r,) + 1, if q < u, - 2, decompose G,
with parameters u, - q, a,, a3. If q > a3, consider Gf which is a graph
obtained by adding to Gi q - a3 of the vertices in T(ri). S(Gl) > 2 and by
Lemma M may be decomposed with parameters a,, u2.
Assume, then, that the remaining possibility holds, where a, = a, = a3 =
q + 1. Let rj have neighbours in T(ri) and let A i = {rir rj} U T(ri). In G\P I
all components are of order >5 and by Lemma M it can be decomposedwith
parameters a,, a3.
We may put the conclusion of the above paragraph in the form
Vm>i>l,
3 1 <j # i < m 3 IT(r,)\T(r,)l
< 2,
(4)
in which case we say that ri hits rj. In what follows Ti stands for T(ri). We
want to show that there are 4 distinct indices m > i,, i,, j,, j, > 1 so that ri,
hits rj,, ri, hits rj,. By (4) this is not the case only if there is a m > t > 1 so
that all ri (m > i # > 1) hit rI and only rt. Let r, hit rs. So
t
Let ri have a neighbour in r,\r,. Since ri hits rI but does not hit rs it follows
that Ir,\r, I = 2 and ri is a neighbour of both vertices in r,\r,. It also follows
that Ir,\r,i = 2 and ri is a neighbour of exactly one vertex in r,\r,.
But
since the vertices in (r,\r,) U (r,\r,)
all have valence 3 (by (3)) this is
impossible.
So we have 4 distinct indices 1 < a, /3, y, 6 < m so that
Now represent the ats as
ai
=fis
+ gi(S
-
1) + hi
(i= 1,2,3),
where 2 fi = f - 2, f being the number of s:s which are =s, and C gi = g =
m -f = number of sI)s which are =s - 1, C hi = 2s, hi > 0. (Iff = I, change
the roles of s and s - 1.)
22
NATHAN LINIAL
We assign now stars to classesas dictated by these parameters, namely,f,
s-starsto A,, etc. Let us say that h r, h, < s/2. Assign S, to A r and S, to A,,
only S,, S, are unassignedyet. Now by (5) we transfer h, vertices of L, to
A i and h, vertices of L, to A,. The rests of S,, S, are assignedto A, to
complete the decomposition.
If h, , h, > s/2, assign S,, S, to A, and transfer s - h, , s - h, vertices
from S,, S, respectively to A,. The remains of S,, S, are assignedto A,,
A z, respectively.
The only case which needs settling yet is the one where s, < 3. Let us say
that we have a Si’s equal 2 and /3 of them equal 3. It is easy to check that if
a > 4 and /3 > 2 then a decomposition exists regardless of the values of a,,
a,, a3.
So we may assumea < 3 or /I < 2. The casesare
a=0
(a,, a,, a3) = (1, 1, 1) or (0, 1,2) or (2,2, 2) mod 3,
a=1
(a,, a,, a3) = (0, 1, 1) or (1,2,2) mod 3,
a=2
(a,, u,, a3> = (2, 1, 1) mod 3,
a=3
(a,, a,, a,) = (1, 1, 1) mod 3,
p=o
(a,,u,,a,)-(0,
1, l)mod2,
p=1
(u,,u,,u,)-(1,
1, l)mod2,
of any of their permutations.
If /I < 1 we can find three 2-stars which can be transformed into two 3stars, taking care of ,8 < 1. If a < 3 we can find two neighbouring 3-stars and
transform them into a 4-star and a 2-star, or else transform four 3-stars into
two 4-stars and two 2-stars. It is a routine check to validate that the decomposition is achieved in any of these cases. 1
Together with the conjecture discussedin the present paper Frank made in
[3] another conjecture, later proved by Lovisz [51 and Gyori [4]:
THEOREM LG. A graph G = (V, E) of order >k + 1 is k-connected iff
for any k integers a, ,..., uk > 1 and any k distinct vertices x, ,..., xk E V, it is
possible to decompose V into A ,,..., A, so that lAil = ui, xi E Ai, (Ai) is
connected (i = l,..., k).
This brings to mind the idea that one should try to prove a stronger
conjecture than the one discussed in the present paper in which not only
a, ,..., ak are specified but also some vertices x, ,..., xk in a manner similar to
Theorem LG. However, even for the case a,= .a- =ak-l =2 the
specification x1 ,..., xk already implies k-connectivity as Theorem 3 shows.
The harder part of the theorem is contained in Theorem LG but it seems
GRAPHDECOMPOSITIONS
23
worth mentioning as it supplies an independent characterization of kconnectivity.
THEOREM 3. A graph G = (V, E) of order >2k - 1 is k-connected iff for
every set {x, ,..., XJ c V, there is a matching of x, ,..., xk-, within G\{xk} so
that the vertices which are not in the matching span a connected subgraph of
G.
Proof: The crucial step in the proof is an application of alternating
paths, a method which is fundamental in matching theory. See Berge 11,
Chap. 8 ] for several examples of this method.
We assumeG to be k-connected, and start by showing that it is possible to
match T= {x, ,..., xk- i} within G\xk. To show this we employ Hall’s
theorem 11,p. 1341. Let X= {xi ..... xi}, and for S c T let N(S) be the set of
those vertices in v\xk which have a neighbour in S. If T cannot be matched
within G\xk, then, by Hall’s theorem, /N(S)1 < 1SI for some S z T. But then
the set W = N(S) U (x\S) separates S from the rest of the vertices in V.
Note that the sets S and W do not exhaust all of V, because together they
contain at most 2k - 2 vertices whereas j VI > 2k - 1. Therefore W
disconnects G, but this is impossible, since
/ WI = IN(
+ 1X1- ISI < 1X1= k.
Among all sets Y that can be matched with T in G\x, we choose one for
which the component of x,, in G\(Tu Y) contains as many vertices as
possible. Assume Y = (y, ,..., all- ,) and [xi,yi] E E for i = l,..., k - 1. If
G\(TU Y) is connected, then the proof is finished, so we assumethat it is
disconnected.
Let C, ,..., C, be the components of G\(Tu Y) and let A, be the vertex set
of Cj. We assumethat r > 2, xk E A,, and that iA, I is as large as possible.
First we note that E(A , , Y) # 0, since otherwise T separatesA, from Y and
therefore from Y U A z U ... WA,., although jTI=k1. Let Y,#0 be the
set of those vertices in Y which have a neighbour in A,. Define
S = (x E T 1There is a sequencex,, ,..., x,, = x of
distinct vertices in T (I > 1) so that J,, E Y, and
(6)
[x~~~,~v~,,~] E E for i = l..... 1- 11.
We show that
E(S,Ai)=0
for
i = 2,..., r.
(7)
Suppose on the contrary that for somex E S there is a y E Ur=2 Ai such
that [X,JJ] E E. Let I, ,,..., .Y,,,= x be a sequenceas in the definition (6). We
24
NATHAN
LINIAL
define Y’ = Y\y,, U y, and show that T can be matched with Y’ in G\xL.
For i & {a, ,..., a,) we leave xi and yi matched. For j = l,..., I- 1 we match
is
xaj with ynj+ i. Note that (xQj, yaj+,] E E by definition (6); x=x,,
matched with y. However, the component of G\(T U Y’) which contains xk
includes A, U y,, , and therefore contains more vertices than the component
C, of G\(TU Y). This contradicts the maximality of (A, 1 and proves (7).
Denote S’ = { yi (xi E S}. We show that (T\S) U S’ separatesS U A, from
(Y\S’)U
UT=, Ai. Since \(T\S) U S’ ] = k - 1 this is a contradiction which
proves the “only if’ part of the theorem. Consider A, first: evidently,
E(A,,Ai)=0fori=2
,..., r. Also E(A,, r\S’) = 0, since Y, 5 S’. As for S,
we have (7). By definition of S we have E(S, Y) = E(S, S’) and this part of
Theorem 3 is proven.
The “if’ part of the theorem is proved as follows: Suppose S c V is such
that 1S I= k - 1 and G\S is disconnected. Let A i ,..., A, (r > 2) be the vertex
sets of the components of G\S. Suppose first that IAil < k - 1 for some i,
and let U be a subset of S having k - 1 - 1Ail vertices. Define xi ,..., xk-, to
be the vertices in Ai U U. Also, let xk be a vertex in S\U. No vertex outside
Ai U S is adjacent to a vertex of A,. Therefore, at most IS\(UU xJ =
\Ai / - 1 vertices in V\(A, U UU xk) may have a neighbour in A i. Thus it is
impossible to match AiU U= {xi ,..., xkAl)
within G\xk.
We may assume,then, that IAil > k for 1 < i < r. Now let x, ,.,., xk-, be
the vertices of S, and xk a vertex not in S. From the assumption that every
component of G\S has >k vertices it follows that for every matching of S in
G (if any), the remaining vertices span a disconnected subgraph of G, a
contradiction. I
Let us show now that the conjecture holds for the casea, = ... = ak- i = 2.
This case is of course a problem on the existence of matchings as was also
noted by Frank and Maurer.
THEOREM
4. Let G be a connected graph of order >2k with 6(G) > k.
Then there is a matching [xi, yi] (i = l,..., k - 1)) so that the graph
G\({xJi=
l,..., - l}U{yili=
l,..., k - 1) ) has no isolated vertices.
Proof. That G contains a (k - 1) matching is known (see, e.g., [2,
Theorem 2.4.21). Consider a matching [xi, yi] (i = l,..., k - 1) for which
G\({x, / i = l,..., k - 1} U { y I i = l,..., k - 1 } has as few isolated vertices as
possible. Let p be an isolated vertex in this subgraph. Identify v\(( p} U
{xi 1i = l,..., k - 1 / U { yi 1i = l,..., k - 1)) to a single vertex q. Let H be the
resulting graph with vertex set {p, q} U {xi ) i = l,..., k - 1) U { yi I i = l,...,
k - 1}, and E = E(H). If we can find a perfect matching in H we can translate this back into a (k - 1) matching in G with fewer isolated vertices
among the vertices which are not in the matching.
25
GRAPH DECOMPOSITIONS
We prove that H has a perfect matching by contradiction. If [p, xi] E E,
[q, yi] E E for some k - 1 > i > 1, then a perfect matching is obtained by
1. Notice that xi, yi
matching [P, Xi], [q, Yi], and [xj,yj], k- 1 >j#i>
play exactly the same roles so whenever an assumption on xi, yi can be
made without loss of generality we will make it with no further comment.
We want to show that for k - 1 > i > 1 either [q, xi], [q,yi] E E or [q, xi],
[q, yi] 6! E. Assume that [q, xi] & E, [q,vi]
E E. By a previous remark we
may assume [p, xi] @E. Now d,(p) > k and [q, xi] 6?E implies that
dH(xi) > k. Therefore both p and xi have at least k - 1 neighbours among the
2k - 4 vertices in U({xj, yj) / k - I > j # i > 1). This implies that for some
j # i, [xi, xi], [p, yj] E E. But now we have the perfect matching [p, yj],
]xi, xj], [q, yi], and [xl, y,] (k - 1 > t f i, j > l), a contradiction.
It follows that there exists a subset I c {l,..., k - 1) so that for i t? I,
[q, xi], [q, y,] 4 E and for i E I, [q, xi], [q, yi] E E. By what was said before,
i E I implies [p, xi], [p, yi] @E. Since G is connected there have to be s E I,
t & I so that [ ys, y,] E E. We repeat a previous argument to conclude that
there is an index j # t so that {xt,p}
can be matched with (xj, yj). Now j
cannot belong to I and in particular j # s. Let us say that [x1, xj] E E,
fp,-vj]
E E. Match these pairs and also [ ys, y,], [q, xs], and [x,, y,] (k - 1 >
Y# s, t,j > 1) for a perfect matching. 1
Note added in proof.
(198 1). 263-273).
Theorem
3 was independently
proved
by E. Gyiiri
(Combinatorics
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